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3 years ago

How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?

Physics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

$W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J$

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

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A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

8 0

3 years ago

It is found that the most probable speed of molecules in a gas at equilibrium temperature

kaheart [24]

Answer:

$\frac{T_2}{T_1} = 1$

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

$v = \sqrt{\frac{3RT}{M} }$

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

T = Equilibrium Temperature

M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

$v = \sqrt{\frac{3RT_1}{M} }$

For T = T₂ :

$v = \sqrt{\frac{3RT_2}{M} }$

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

$\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1$

8 0

3 years ago

Which of the following is most appropriate to follow motion in one dimension?

Lyrx [107]

The linear scale is applicable only as it moves in one dimension. From the word "linear" it means it deals with one equation only. Unlike the other options, the dimensions are many because it involves 2 or more variables for its equation.

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3 years ago

PLEASE HELP AS FAT WILL GIVE BRAILIEST TO MOST SPECIFIC AND CORRECT ANSWER

Bas_tet [7]

Answer:

Time

Explanation:

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3 years ago